A ball of mass 10 kg moving with a velocity 10√3 m s⁻¹ along X-axis, hits another ball of mass 20 kg which is at rest. After collision, the first ball comes to rest and the second one disintegrates into two equal pieces. One of the pieces starts moving along Y-axis at a speed of 10 m/s. The second piece starts moving at a speed of 20 m/s at an angle θ (degree) with respect to the X-axis.
The configuration of pieces after collision is shown in the figure.
The value of θ to the nearest integer is ____________.

Solution

Physics sir · 4 steps

Nothing outside acts on the balls, so total momentum is conserved. Balance it along X and along Y.

1. Before collision. Only the 10 kg ball moves. The total momentum is 10 × 10√3 = 100√3 kg m s⁻¹ along X, and zero along Y.
2. After collision. The first ball stops. The 20 kg ball splits into two pieces of 10 kg each. The piece moving up the Y-axis carries 10 × 10 = 100 kg m s⁻¹. The other piece carries 10 × 20 = 200 kg m s⁻¹ at θ below the X-axis.

   X: 200 cos θ = 100√3 Y: 100 - 200 sin θ = 0

3. Solve. From X, cos θ = √3/2. From Y, sin θ = 1/2. Both give the same angle, which confirms the working.
4. So. θ = 30°, so the answer is 30.

The common slip: 64. Taking the moving piece as 20 kg instead of 10 kg gives cos θ = √3/4 and θ ≈ 64°. Each half of the 20 kg ball has a mass of only 10 kg.

The idea to keep. In any break-up or collision with no external force, write momentum conservation separately along each axis. Always use the mass of each piece, not the mass of the parent body.

Answer from the NTA final answer key. Solution by PrepWiser.